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Helen [10]
2 years ago
12

If environmental triggers cannot be avoided or modified, they may make it difficult to extinguish a behavior. please select the

best answer from the choices provided t f
Physics
1 answer:
Free_Kalibri [48]2 years ago
6 0

The answer is True . An external or internal trigger can occur Environment-based factors serve as external triggers. They are detected by our minds as well as by our five senses.

Internal triggers are unrelated to anything outside of us and are caused by our ideas and feelings. A stimulus in physiology is a perceptible alteration in the internal or external surroundings of an organism's physical or chemical composition. Sensitivity is the capacity of an organism or organ to recognize external stimuli and respond accordingly (excitability). The function of that behavior is the explanation for why a behavior persists in Applied Behavior Analysis (ABA). The child receives reinforcement from these activities. The behavior will stop and be replaced by a new behavior if it ceases to serve that purpose.

Learn more about to environmental triggers visit here:

brainly.com/question/28349211

#SPJ4

You might be interested in
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the
sashaice [31]

Answer:

52.47706 mph

54.5 mph

Explanation:

The average speed is given by

V_{av}=\dfrac{Distance}{Time}

V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph

Julie's average speed on the way to Grandmother's house is 52.47706 mph

V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph

Average speed on the return trip is 54.5 mph

4 0
4 years ago
Scientists are making plans to put a probe in orbit around Earth. They want the probe to enter the orbit shown below.
iris [78.8K]

An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.

<h3>What is an orbit?</h3>

An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.

In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.

Read more on orbit here: brainly.com/question/18496962

#SPJ1

6 0
3 years ago
Read 2 more answers
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

7 0
3 years ago
Find the weight of the following mass; <br> (a) 220g <br>please show the formula​
Marrrta [24]

Explanation:

weight=mass*gravity

w=mg

first change 220g to kg like this

1kg=1000g

?=220g

1kg*220g/1000kg=220/1000g

mass =0.22kg

now we can continue

w=0.22kg*10m/s^2 or w=0.22kg*10N/kg

=2.2N =2.2N

hope it helps you

5 0
4 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0
4 years ago
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